Induction.

The induced currents curl around the cylinder axis, since that is the direction of the induced electric field. Divide the cylinder into infinitesimally thin loops stacked together radially, and sum the currents in all of them. Ignore self-induction effects.

Flux Linkage in Loop of Radius $r$

$\lambda = B(t) \, \pi r^2 = B_0 t \, \pi r^2$

Voltage Induced in Loop: $V = \frac{d \lambda}{dt} = \pi B_0 \, r^2$

Loop Resistance: $R = \frac{\rho \, l}{A} = \frac{(1/\sigma) \, 2 \pi r}{h \, dr}$

Loop Current:

$dI = \frac{V}{R} = \frac{\pi \, B_0 \, h \, \sigma \, r^2 dr}{2 \pi \, r} = \frac{ \, B_0 \, h \, \sigma \, r dr}{2}$

Total Current:

$I = \frac{ \, B_0 \, \sigma \, h}{2} \int_a^b r \,dr = \frac{ \, B_0 \, \sigma \, h}{4} (b^2 - a^2)$