Confused about frequency and bandwidth in BP filter?
To answer your last question first, you can't have a narrow bandwidth filter around 1MHz and yet still have a fast rise time. If you think about the spectrum of a square wave, it has frequency components extending to infinity. The higher frequency components contribute to the 'squareing up' of the signal and sharpening of the edges. e.g. see http://mathworld.wolfram.com/FourierSeriesSquareWave.html  Having a narrow band around 1MHz means your signal will come out looking like a 1MHz sine wave.

With that in mind you have to design a bandpass filter that does not attenuate your 1MHz fundamental frequency too much, yet includes high enough frequencies to give the desired rise time. Following your formula, 0.34 = rise time * bandwidth, you have calculated a bandwidth of 34MHz is required. The next step is to consider bandwidth = high cutoff freq. - low cutoff freq. You want the low cutoff to be less than 1MHz. Let's choose 500kHz. Thus the high cutoff would be 34.5Mhz and the centre frequency 17.25Mhz.

To get rid of the most noise, the filter should have a steep rolloff, e.g. the two filters mentioned in the comments. This means your low cutoff frequency can be very close to 1MHz without too much attenuation, and in the higher end of the spectrum rolls off very quickly after the high cutoff frequency, reducing high frequency noise.