closed loop gain and bandwith calculation
There is a fairly simple graphical method that can be used to solve this.  It's derived from the expression for the closed-loop gain for a negative feedback system:
$$
\text{Gain} = \frac{A}{1+AB}
$$
where A is the open-loop gain and B is the feedback factor. We can see from this that for large loop gain AB
$$
\text{Gain} \approx \frac{1}{B}
$$
In this example, A is the open-loop gain of the op amp. The feedback factor B is the feedback signal divided by the output signal with the input at zero, which in this case is
$$
B = \frac{R_1}{R_1+R_f}
$$
where \$R_f\$ is the equivalent resistance of the switched resistor network.  We have here \$R_1 = 10\text{k}\Omega\$ and \$R_f = 14.13\text{k}\Omega\$ for the switch in position 7. (The 1pF cap can be neglected.)  

To use this method, plot \$|A| \text{ and } \frac{1}{B}\$ on a Bode plot:  The LF351 is dominant-pole compensated, so the plot of \$|A|\$ is a line sloping down at -20dB\decade and crossing zero dB at 4MHz, the gain-bandwidth product.  Since B = 0.4144, the plot of \$\frac{1}{B}\$ is a horizontal line at 7.65dB. 

The frequency where these two lines cross is the closed-loop bandwidth.  In this case, they cross at 1.66MHz.  The \$\frac{1}{B}\$ line to the left of the bandwidth frequency and the \$|A|\$ line to the right are assymptotes of the closed loop frequency response curve.

This method is valid for inverting and non-inverting configurations, BUT for non-inverting, as in this case, the input signal "seen by" the summing point is attenuated by
$$
\frac{R_f}{R_1+R_f} = 1-B
$$
so the overall closed-loop gain is \$\frac{(1-B)}{B}\$ rather than \$\frac{1}{B}\$. This has the effect of shifting down the frequency response curve by \$20\text{ log}(1-B)\text{ dB}\$, which is -4.65dB in this case, so we have closed-loop gain = 3dB.