Solution :
given $V= 50\;V$
$I= 12\;A$
n$= 30\%$
Input power = VI
$\qquad= 50 \times 12 =600 walt$
Since $n=30\%$ , the energy dissipated as heat
$P= \large\frac{70}{100}$
$\quad= \large\frac{70}{100} $$ \times 600$
$\qquad= 420\;w$
$P= I^2 R$
$R= \large\frac{P}{I^2}$
$\quad= \large\frac{420}{12 \times 12}$
$\quad= 2.9 \Omega$
or $ 3\;\Omega$ (approx)