Answer
$k=\dfrac{2}{27}$
Work Step by Step
Given: $\int_{1}^{2}\int_{0}^{3} kx^2y dx dy=1$
Consider $\int_{1}^{2}\int_{0}^{3} kx^2y dx dy=k \int_{1}^{2}y[\dfrac{x^3}{3}]_{0}^{3} dy$
This implies that
$V=9k \int_{1}^{2} (y) dy=9k[\dfrac{y^2}{2}]_1^2$
Then, we have $9k[2-\dfrac{1}{2}]=9k(\dfrac{3}{2}^=\dfrac{27}{2}k$
Hence, $\dfrac{27}{2}k=1 \implies k=\dfrac{2}{27}$
