= Solution
For an oriented piecewise <smooth surface> $S$ and a <continuously differentiable> <vector field>, <Stokes theorem> states
$$
\int_S(\nabla\times B)\cdot n\,dS
=\oint_{\partial S}B\cdot dr,
$$
where the <boundary orientation> is induced by $n$ through the <right-hand rule>. Choose the upward orientation of the <elliptic paraboloid>. Its band has <cylindrical radius> $1/2\leq r\leq1$, and a parametrization is
$$
X(r,\varphi)=(r\cos\varphi,r\sin\varphi,r^2),
\qquad 0\leq\varphi<2\pi.
$$
Its <oriented surface element> and scalar area element are
$$
\boxed{d\mathbf S=(X_r\times X_\varphi)\,dr\,d\varphi
=(-2r^2\cos\varphi,-2r^2\sin\varphi,r)\,dr\,d\varphi,}
$$
$$
dS=r\sqrt{1+4r^2}\,dr\,d\varphi.
$$
The requested sketch shows the open band, not a capped solid.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ia/paper-3-paraboloid.png]
{title=Upward-oriented annular paraboloid band, with counterclockwise outer boundary and clockwise inner boundary}
For the specified <vector field>, direct <differentiation> gives $\nabla\times B=(0,0,3x^2+3y^2)$. Hence
$$
I=\int_0^{2\pi}\int_{1/2}^1 3r^3\,dr\,d\varphi
=\boxed{\frac{45\pi}{32}}.
$$
To check this by <Stokes theorem>, parametrize a counterclockwise circle of radius $\rho$ at height $\rho^2$ by $r_\rho(\varphi)=(\rho\cos\varphi,\rho\sin\varphi,\rho^2)$. Along it $dz=0$, and
$$
B\cdot dr_\rho
=\rho^4(\sin^4\varphi+\cos^4\varphi)\,d\varphi.
$$
Using $\int_0^{2\pi}\sin^4\varphi\,d\varphi =\int_0^{2\pi}\cos^4\varphi\,d\varphi=3\pi/4$ gives $3\pi\rho^4/2$. The induced outer circle is counterclockwise as viewed from above, and the inner circle is clockwise. Therefore
$$
I=\oint_{C_1}B\cdot dr-\oint_{C_{1/2}}B\cdot dr
=\frac{3\pi}{2}\left(1-\frac1{16}\right)
=\boxed{\frac{45\pi}{32}}.
$$
This verifies the direct <surface integral> and the <Stokes flux through an annular paraboloid>. Reversing the chosen surface orientation reverses the final sign.
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