= Solution
\b[In the stated model, the limiting thicknesses are about $0.163\ \mathrm m$ of lead or $0.780\ \mathrm m$ of concrete.] Use the unshielded <equivalent dose> rate from part (c), $\dot H_0=556.815\ \mathrm{\mu Sv\,h^{-1}}$. Let $x=\mu t$, where $\mu$ is the supplied total <linear attenuation coefficient>. The stipulated <radiation buildup factor> gives the requirement
$$
\dot H_0(1+x)e^{-x}<3\ \mathrm{\mu Sv\,h^{-1}},
\qquad
(1+x)e^{-x}<\frac{3}{556.815}=0.00538778.
$$
The left side decreases for $x>0$ because its derivative is $-xe^{-x}$. Solving the equality gives a unique threshold,
$$
x_*=7.34532.
$$
Thus, with the source-to-observation distance kept at one metre,
$$
\boxed{t_{\rm Pb}>\frac{x_*}{45.2}=0.162507\ \mathrm m,
\qquad
t_{\rm concrete}>\frac{x_*}{9.42}=0.779758\ \mathrm m.}
$$
For example, rounding upward to $0.17\ \mathrm m$ of <lead> or $0.80\ \mathrm m$ of concrete satisfies the stipulated inequality. Equality at the threshold is not sufficient for the strictly smaller target.
A compact surrounding shield favours <lead>; concrete trades much greater thickness for a common structural material. The densities permit a useful planar comparison: the threshold areal masses $\rho t$ are about $1.84\times10^3$ and $1.87\times10^3\ \mathrm{kg\,m^{-2}}$, respectively. \b[The thinner shield is not proportionately lighter per equal projected area.] Total enclosure <mass> also depends on geometry.
The idealized outline is continuous coverage around the source with the required material path length, including the closure; a direct unshielded line through a gap would invalidate this attenuation estimate. The calculation assumes the quoted observation distance, the fixed effective photon coefficients and the prescribed <radiation buildup factor>. It is not based on the more optimistic uncollided <exponential attenuation> alone.
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