= Solution
Orient $C$ counterclockwise around the solid body and let $\mathbf n$ point from the solid into the fluid. The <pressure> <force> is $-\int_Cp\mathbf n\,ds$. Since $\mathbf n\,ds=(dy,-dx)$,
$$
F_x-iF_y=-i\oint_Cp\,d\overline z.
$$
For steady <irrotational flow>, <Bernoulli's equation> gives $p=p_0-\rho|w'|^2/2$. The constant term integrates to zero. The boundary is a <streamline>, so $dw=w'\,dz$ is real there, giving $w'\,dz=\overline{w'}\,d\overline z$. Therefore
$$
F_x-iF_y=\frac{i\rho}{2}\oint_C|w'|^2\,d\overline z
=\frac{i\rho}{2}\oint_C(w')^2\,dz,
$$
which is <Blasius force theorem> with the specified orientation.
The prescribed <stream function> is the imaginary part of $w_0(z)=\alpha(z-c)^3$. The zero-circulation solution from the <Milne-Thomson circle theorem> is
$$
w(z)=w_0(z)+\overline{w_0(a^2/\overline z)}
=\alpha\bigl[(z-c)^3+(a^2/z-c)^3\bigr].
$$
On $|z|=a$ the two terms are conjugates, so the boundary <stream function> is constant. The added term has no singularities in the exterior and does not change the prescribed growing far-field part. Additive constants are immaterial.
Differentiating gives
$$
w'=3\alpha\bigl[z^2-2cz+c^2-c^2a^2z^{-2}+2ca^4z^{-3}-a^6z^{-4}\bigr].
$$
The coefficient of $z^{-1}$ in $(w')^2$ is $36\alpha^2ca^2(a^2+c^2)$. The <residue theorem> and <Blasius force theorem> give
$$
\boxed{F_x=-36\pi\rho\alpha^2ca^2(a^2+c^2),\qquad F_y=0.}
$$
The zero-circulation convention is relevant: if an independently imposed <circulation> $\Gamma$ is allowed, adding $\Gamma\log z/(2\pi i)$ changes $F_y$ to $-3\rho\alpha\Gamma c^2$. Thus the claimed zero transverse <force> belongs to the standard circle-theorem construction, or to a single-valued exterior <complex potential of a two-dimensional potential flow>, which excludes that extra circulation.
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