= Solution
In a long wire away from its ends, the steady <temperature> is independent of axial position and azimuth. The <heat equation> reduces to
$$
\frac1\rho\frac{d}{d\rho}\left(\rho\frac{dT}{d\rho}\right)=-\frac{H}{\kappa}.
$$
Multiplying by $\rho$ and integrating gives
$$
\rho T'=-\frac{H\rho^2}{2\kappa}+C_1,
\qquad
T=-\frac{H\rho^2}{4\kappa}+C_1\ln\rho+C_2.
$$
A finite <temperature> and nonsingular flux at the axis require $C_1=0$. Applying $T(a)=T_0$ determines $C_2=T_0+Ha^2/(4\kappa)$. Therefore
$$
\boxed{T(\rho)=T_0+\frac{H}{4\kappa}(a^2-\rho^2)
=T_0+\frac{I^2}{4\pi^2a^4\kappa\sigma}(a^2-\rho^2).}
$$
This is the <steady temperature of a uniformly heated cylindrical wire>. Its hottest point is at the centre, with $T(0)-T_0=I^2/(4\pi^2a^2\kappa\sigma)$. The outward <heat flux> is $J_Q(\rho)=-\kappa T'(\rho)=H\rho/2$, so the <power> crossing a cylindrical surface per unit length, $2\pi\rho J_Q(\rho)=H\pi\rho^2$, equals the heating generated inside it.
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