Answer
Please see the work below.
Work Step by Step
As $P_{out}=P_{in}-P$
We plug in the known values to obtain:
$P_{out}=750-(750*.35)= 488 MW$
We know that
$\Delta T=\frac{P_{out}}{mc}$
$\implies \Delta T=\frac{488 \times 10^6}{1.1\times 10^5\times 4184}=1.1 K$
As the temperature of river increases by 1.1 K, so the standard can be achieved by using river water for cooling.
