Use the first law of thermodynamics in rate form

The appliance supplies heat to the system at the rate $\dot Q = 6000\ \text{J/min} = \frac{6000}{60} = 100\ \text{J/s} = 100\ \text{W}.$

The system delivers power $P = 90\ \text{W}.$

So the rate of increase of internal energy is $\frac{dU}{dt} = \dot Q - P = 100 - 90 = 10\ \text{W}.$

Find the time required for the given increase in internal energy

Required increase: $\Delta U = 2.5 \times 10^3\ \text{J}.$

Since $\Delta U = \left(\frac{dU}{dt}\right)t,$ we get $t = \frac{\Delta U}{dU/dt} = \frac{2.5\times 10^3}{10} = 2.5\times 10^2\ \text{s}.$

Match with options

$t = 2.5 \times 10^2\ \text{s}.$

This corresponds to Option A.