Increasing frequency response instead of gain for a Cascode circuit
When you analyze such a circuit for the first time you cannot avoid drawing the small signal equivalent circuit, so start with that.

Next step is to identify the corner frequencies. The easiest one is of course \$R_l C_l\$ or more accurately: \$R_l *(C_l + C_{ds,Mcasc})\$

Another one is at the input, the \$C_{gd,Min}\$ times the source resistance. But in order to say anything about the influence of this corner frequency we need to know the output impedance of the source driving. That's where \$R_{sig}=0\$ comes in, meaning we can ignore this pole (You know why ?)

Also note how the bode plot only shows a 1st order low-pass characteristic, it means there's only one corner frequency.

So the conclusion is that there's only one corner frequency to consider which is \$R_l *(C_l + C_{ds,Mcasc})\$

Now what happens if for example \$C_l\$ doubles in value ? The RC product becomes larger. But is RC in the nominator or denominator of the formula for a corner frequency ? So what happens to the value of that corner frequency when \$C_l\$ is doubled ? So what must be done to make the corner frequency higher (increasing the bandwidth) ?