Hi Digambar,

Given:

T1 = 2000 K

T2 is unknown.

T3 = 200 K

As carnot engines are connected in series work done will be equal in both: 

W1 = Q2 - Q3

W2 = Q1 - Q2

Since, W1 = W2

Q1 - Q2 = Q2 - Q3

Q2 = (Q1 + Q3) / 2

In carnot cycle, Q ∝ T, hence

T2 = (T1 + T3) / 2

T2 = (2000+200) / 2

T2 = 1100K.

Efficiency of 1st engine -

nA = (Q1 - Q2) / Q1

nA = (T1 - T2) / T1

nA = (2000-1100) / 2000

nA = 0.45

Hope it helps.