= Solution
A $90^\circ$ <gravitational scattering angle> has $\cot(\phi/2)=1$. Thus
$$
\boxed{b=\frac{GM_J}{v_0^2}
=\frac{(6.6743\times10^{-11})(2\times10^{27})}{(10^4)^2}\,\mathrm m
\simeq1.33\times10^9\,\mathrm m.}
$$
The denominator uses the squared <speed> in $\mathrm{m\,s^{-1}}$, not its numerical value in $\mathrm{km\,s^{-1}}$. From <11/b>, the corresponding <periapsis distance> is $r_{\min}=(\sqrt2-1)b\simeq5.53\times10^8\,\mathrm m$; this differs substantially from the <impact parameter> because of gravitational focusing.
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