Answer
$T_f = 320.25 K$
Work Step by Step
From equation $Q = mc(T_f - T_i)$
In equilibrium state, final temperature, $T_f$ is the same in both blocks. So
$m_{Cu}c_{Cu}(T_f - T_{i(Cu)}) =m_{Pb}c_{Pb}(T_f - T_{i(Pb)}) $
Here we solve for $T_f$
$T_f = \frac{m_{Cu}c_{Cu}T_{i(Cu)} + m_{Pb}c_{Pb} T_{i(Pb)}}{m_{Cu}c_{Cu} + m_{Pb}c_{Pb}} $
$T_f = \frac{(50 g)(386 J/kg.K)(400K) + (100 g)(128 J/kg.K)(200K)}{(50 g)(386 J/kg.K) + (100 g) (128 J/kg.K)} $
$T_f = 320.25 K$
