60 liters of 40% HNO3 solution
The concentration of HNO3 present in the solution is 40 liters in 100 liters for 40% HNO3
The concentration of HNO3 in 60 liters is
$$\frac{60\cdot40}{100}=24\ liters$$
Let the volume of pure undiluted HNO3 = x
Volume of pure HNO3 to add in order to increase concentration =>
24 + x = 50% of (60+x)
24 + x = (1/2) (60+x)
2(24+x) = 60+x
48 + 2x = 60 + x
2x - x = 60 - 48
x = 12 liters of pure HNO3 must be added in order to have a concentration of HNO3 solution.

Answer = option A