$V_{sec}$=50v(ms)

So, $V_{m}=50\sqrt 2$=70.71V

$I_{m}=\frac{V_{m}}{R_{L}+R_{f}}=\frac{70.71}{980+20}$=70.71 mA

(i) Average Load Current

$I_{dc}=I_{avg}=\frac{2I_{m}}{\pi}$=45.01 mA..........(i)

ii)$I_{rms}$ load current

$I_{rms}=\frac{I_{m}}{\sqrt 2}$=49.99mA....................................(ii)

iii) $v_{dc}=R_{L}-I_{dc}$

=(980)(45.01)mA=44.10V ....................................(iii)

iv)PIV=$2v_{m}$

$2\times70.71v$=141.42V ..............................(iv)

v)Efficiency $\eta=\frac{0.812}{1+(R_{f}/R_{L})}\times100%$

$\eta=\frac{0.812}{1+(20/980)}$=79.57%