= Solution
Substitution in the <moment of inertia> gives
$$
\boxed{I=\frac12\pi(8\times10^3)(0.01)(1)^4=40\pi\simeq126\,\mathrm{kg\,m^2}.}
$$
For independent errors interpreted as <standard uncertainties>, logarithmic differentiation and <propagation of uncertainty> give
$$
\left(\frac{u_I}{I}\right)^2=\left(\frac{u_\rho}{\rho}\right)^2+\left(\frac{u_s}{s}\right)^2+16\left(\frac{u_R}{R}\right)^2.
$$
Consequently
$$
\boxed{\frac{u_R}{R}\leq\frac14\sqrt{0.025^2-0.01^2-0.01^2}=0.00515,\qquad u_R\leq5.15\,\mathrm{mm}.}
$$
This is a $0.515\%$ statistical precision requirement. If “accuracy” instead means hard bounds whose worst cases must be guaranteed, first-order absolute errors add: $0.01+0.01+4\epsilon_R\leq0.025$, giving $\boxed{\epsilon_R\leq0.125\%,\quad |\delta R|\leq1.25\,\mathrm{mm}}$ to first order. For an exact symmetric hard bound about the nominal values, the upper extreme is slightly more restrictive: $(1.01)^2(1+\epsilon_R)^4\leq1.025$ gives $\epsilon_R\leq0.1199\%$, or $1.20\,\mathrm{mm}$. The question does not distinguish a statistical uncertainty from a worst-case tolerance, so the convention matters.
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