Answer:
current in loops is 52.73 μA
Explanation:
given data
side of square a = b  = 2.40 cm = 0.024 m
resistance R = 1.20×10^−2 Ω
edge of the loop c  = 1.20 cm = 0.012 m
rate of current = 120 A/s
to find out
current in the loop
solution
we know current formula that is
current = voltage / resistance    .................a
so current = 1/R × d∅/dt
and we know here that 
flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c)    ...............b
so 
d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt       ...........c
so from equation a we get here current 
current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt 
current = ( 4π× ×0.024 / 2π(1.20×
×0.024 / 2π(1.20× ) × ln (0.024 + 0.012/0.012) × 120
) × ln (0.024 + 0.012/0.012) × 120 
solve it and we get current that is 
current = 4 × × 1.09861 × 120
× 1.09861 × 120 
current = 52.73 × A
  A
so here current in loops is 52.73 μA