A field-effect transistor amplifier obeys a constant gain-bandwidth product: raise the gain and the usable bandwidth falls in proportion. The upper cutoff frequency is fixed by how fast the transconductance can charge the total node capacitance.
Write the bandwidth directly as
\[\text{BW} = \frac{g_m}{2\pi C\,|A_v|}.\]
Here $g_m = 0.095\ \text{S}$, $C = 5 \times 10^{-9}\ \text{F}$ and $|A_v| = 30$.
First evaluate the gain-bandwidth factor $g_m/(2\pi C) = 0.095 / (3.14 \times 10^{-8}) \approx 3\ \text{MHz}$. This alone is the largest bandwidth the device could give at unity gain.
Now share that budget over the required gain of 30:
\[\text{BW} = \frac{3 \times 10^{6}}{30} \approx 100\ \text{kHz}.\]
So the amplifier trades its $3\ \text{MHz}$ speed for a factor-of-thirty voltage gain, leaving a $100\ \text{kHz}$ bandwidth. \[\boxed{\text{BW} \approx 100\ \text{kHz}}\]