= Solution
Taking the <Laplace transform> in $t$ and using the initial condition $u(r,0)=u_0$, the <radial heat equation in three dimensions> becomes
$$
sU-u_0=\frac1r(rU)_{rr}.
$$
Put $W=rU$ and $q=\sqrt s$. Then
$$
W''-sW=-u_0r,
$$
so
$$
U(r,s)=\frac{u_0}{s}+\frac{A(s)\sinh(qr)+B(s)\cosh(qr)}r.
$$
Finiteness at the centre forces $B(s)=0$.
The transformed boundary condition is
$$
\frac1kU_r(a,s)=\frac{u_0}{s}-U(a,s)-\frac1{s^2}.
$$
Since
$$
U_r(a,s)=A(s)\frac{qa\cosh(qa)-\sinh(qa)}{a^2},
$$
substitution and cancellation of the two $u_0/s$ terms gives
$$
A(s)\left[\frac{qa\cosh(qa)-\sinh(qa)}{ka^2}+\frac{\sinh(qa)}a\right]
=-\frac1{s^2}.
$$
Hence
$$
A(s)=-\frac{ka^2}{s^2\{qa\cosh(qa)+(ka-1)\sinh(qa)\}},
$$
and the required explicit transform is
$$
\boxed{
U(r,s)=\frac{u_0}{s}
-\frac{ka^2\sinh(r\sqrt s)}
{r s^2\{a\sqrt s\cosh(a\sqrt s)+(ka-1)\sinh(a\sqrt s)\}}
}.
$$
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