Step 1: Rewrite the curve:
$y=\dfrac{x^2}{b}$, so $y=1$ gives $x=\sqrt b$ and $y=4$ gives $x=2\sqrt b$.
Step 2: Area by rectangles:
The area is the rectangle area under $y=4$ minus the area under the curve: use $\int_{\sqrt b}^{2\sqrt b}\left(4-\frac{x^2}{b}\right)dx$ plus the strip $0\le x\le\sqrt b$ between $y=1$ and $y=4$, which is $3\sqrt b$.
Step 3: Compute:
$\int_{\sqrt b}^{2\sqrt b}\left(4-\frac{x^2}b\right)dx=4\sqrt b-\frac{7b\sqrt b}{3b}=4\sqrt b-\frac{7\sqrt b}{3}=\frac{5\sqrt b}{3}$. Adding $3\sqrt b$ gives $\frac{14\sqrt b}3$.
Step 4: Solve:
$\frac{14\sqrt b}3=28$ gives $\sqrt b=6$, so $b=36$.
Final Answer:
Both ways give area 14 sqrt(b)/3 = 28.
\[ \boxed{A} \]