In a fullwave rectifier, the load resistance R_L = 2 kΩ. Each diode has idealized characteristics having slope corresponding of 400 Ω. Voltage applied to each diode is 240 sin 50 t. V. The peak value of current, I_dc is: 

concept:

In Full wave recifier, 

if v(t)= V_m sinwt

then, I_dc =\( \frac {2Im} {\pi}\)

where, I_m = \(\frac{Vm}{R_f + R_S +R_L​} \)

R_f is diode resistance

Rs is source resistance(transformer winding resistance)

RL is load resistance

calculation:

v(t)=240 sin 50 t

V_m = 240 

R_f =400

R_L= 2000

I_m =\(\frac{240}{2000+400}\) 

=\(\frac{240}{2400}\)

=0.1

I_dc = \(\frac{0.2}{\pi}\)

=0.06369

63.69mA

so correct option is 3.