Two beams have the same frequency, and one is much brighter.

Step 1. (a) Every photon has $h\nu$, which depends only on the frequency, so $KE_{\max}$ is the same. A brighter beam brings more photons each second, and each frees one electron, so more electrons flow: a larger current.

Step 2. (b) $I = 2.5 \times 10^{15}\ \text{s}^{-1} \times 1.6 \times 10^{-19}\ \text{C} = 4 \times 10^{-4}\ \text{A} = 0.4\ \text{mA}$.