Bjt Multistage Amplifier
So your noise floor is 100 nanoVolt PeakPeak (about 100/6.2 = 16 nanoVolt RMS, for 1ppm noise crests ). Over 1 Hertz to 200Hertz bandwidth , which we'll assume is -3dB F3dB asymptotic points.

You want a Signal_noise_ratio (SNR) of ??? Not enough information to define that.

So just use some practical achievable parameters.

The 16nanoVolt/200Hertz implies a per_root_Hertz_random_electron_noise_density of 16/sqrt(200) = 16/14 = 1.2 nanoVolts/rtHz RMS noise === 80 ohms Rnoise.

This means the " rbb' " parameter an be 80 ohms at most. In a system design, allocate 40 ohms to that param (you'll want a small-signal switching transistor bipolar, where the low-value rbb' means the base charges can be quickly extracted for quick turnonn). The other 40 ohms is used in GainSetting feedback network as the lowest Resistor of several resistors.

DC_blocking provides the 1Hertz lower frequency corner, used as High Pass Filter.

A capacitor across the larger of the two gain-set resistors, setting a time constant of 1/(200Hz * 6.28) == 1/1260 = 800 microSeconds, is the upper frequecy corner.

Simply go for a modest gain inn the first stage: 10X or 20X or maybe 50X or 100X. Then have a second stage. This eases the amplifier biasing.

You can use classis RIAA vinyl audio playback preamplifier SCHEMATICS for this circuit.

How much total gain? That is not clear, but lets use the 60dB dynamic range as our guide. Also use the 10voltsPP output.

1) That could mean 1,000:1 variation minimum-to-maximum signal input.  Thus the minimum output voltage is 10v/1,000 = 10 milliVolts peakpeak Vout.

Or may mean the ratio of 1,000:1 for minimum_signal/typical_noise; with 100 nanoVolts peakpeak random noise, we might concluded the minimum signal is 1,000X the noise, or 1,000 * 100 nanoVolts peakpeak = 100,000 nanoVolts PP input signal which is 100 microVolts pp input. And you should provide enough gain to boost that 100 microVolts to be the 10 volts output. That would also cause output distortion/clipping//overload if the sensor provides more than 100 microVolts PP.

Without a schematic, we can't pick the resistors, capacitors, and nor can we set the operating point. Probably need +- 15 volts, so can easily support 10 volts peakpeak output voltage swing. This helps preserve linearity, reducing distortion, and allowing better geophysical signal processing to model the oil reservoir.

Notice we've ignored the 4,000 ohm input impdance. That might be used to set the input_DC_blocking_capacitor value, to achieve the 1Hertz High Pass filter.

================================ design of stage #1 (gain = 100x)

Lets let your gain at 100X, using a voltage divider of 39 ohms and 3,900 ohms in series. The 39 ohms goes to Ground and to the 3,900 ohms. The midpoint ties to the emitter of a low-noise NPN; I've suggested you pick a NPN labeled fast-switching small-signal, so the rbb' is small.

Connect the NPN's collector to 10,000 ohm resistor, that connected to +15 volts.

Now we need to buffer and bias this transistor.

Now for the 2nd NPN: connect its collector to +15 volts; you can insert 100 ohms if you want short-circuit protection.

Tie the 2nd base to collector of the first. 

Tie the 2nd emitter to top of the 3,900 ohms.

This gives us the AC_feedback to establish the gain.

We still need to control the operating point.

Given we have +15 volts, for maximum linearity, we'd like +-7 volts swing on the output, which means +-7 volts on the collector of first NPN. Yet the base is sitting about 0.6 volts up from Ground.

What we'll do is divide down the output DC voltage by 10:1, to be 0.7 volts; we'll place a large capacitor on the 0.7 volts, to prevent large AC feedback; and we'll have a final resistor between the 0.7 volts (and the large cap) to base of our first NPN, the resistor sized to drop 0.1 volts while providing base current as needed to have collector current as needed to drop 7 volts across the 10,000 ohm resistor in the collector of the first NPN.

To provide good pulldown for that 2nd NPN (our buffer; emitter follower; common collector), use another 3,900 ohms to a 390 ohms to ground. This provides the DC feedback of about 0.7 volts. 

Your task is 

(1) to compute the resistor from the 3,900:390 DC_feedback divider to the base of NPN #1, to get enough base current to drop 7 volts across the 10,000 ohm collector. Pick this resistor with expectation of only 0.1 volts across it.

(2) compute the large capacitor across the 390 ohm resistor in the DC_feedback, to establish a 1Hertz High Pass response.

(3) compute the small capacitor (we have not yet talked aobut this) across the AC_feedback (3,900:39 ohm) network, specifically across the 3,900 ohm, to establish the 200Hertz low pass response.

Note the amplifier noise floor might be set by the resistor from base of NPN#1 to the DC_feedback divider (top of the large cap).

There are various alternatives to these feedbacks, but that is for another day, once you get simulations running on this.