Answer
(a) $0.800m$
(b) decreased by a factor of 2.
Work Step by Step
(a) We know that
$L=\frac{A\Delta T(K_{Aluminum+K_{St.steel}})}{H}$
We plug in the known values to obtain:
$L=\frac{\pi(\frac{3.50\times 10^{-2}m}{2})^2(98K)(217W/m.K)+(16.3W/m.K)}{27.5J/s}$
$\implies L=0.800m$
(b) We know that
$L\propto \frac{1}{H}$
This relation shows that if the length of the rods is doubled then the rate of heat flow is decreased by a factor of $2$.