$$\text{RHS : } U(\Lambda^{-1}\Lambda'\Lambda)=U(\Lambda^{-1}(1+\delta w)\Lambda)=U(1+(\Lambda^{-1}\delta w\Lambda))$$ $$\Rightarrow I \;+ \frac{i}{2\hbar} (\Lambda^{-1}\delta w\Lambda)_{\rho \sigma}M^{\rho \sigma} \tag{1}$$
$$ (\Lambda^{-1} \delta w \Lambda)_{\rho \sigma}={(\Lambda^{-1}\delta w \Lambda)^{\alpha}}_{\sigma}g_{\alpha \rho}={(\Lambda^{-1})^{\alpha}}_{\mu}{\delta w^{\mu}}_{\nu}{\Lambda^{\nu}}_{\sigma} g_{\alpha \rho} = {\Lambda_{\mu}}^{\alpha}{\delta w^{\mu}}_{\nu}{\Lambda^{\nu}}_{\sigma} g_{\alpha \rho} = {\Lambda^{\mu}}_{\rho}\delta w_{\mu \nu}{\Lambda^{\nu}}_{\sigma} \tag{2}$$
Substitute $(2)$ in $(1)$ and you should get the desired result.
$$\underline{\text{Elaboration}}$$
$$\text{ LHS : }U(\Lambda^{-1})U(\Lambda')U(\Lambda)=U(\Lambda^{-1})(I+\frac{i}{2\hbar}\delta w_{\mu \nu} M^{\mu \nu})U(\Lambda)$$ $$\Rightarrow I+\frac{i}{2\hbar}\delta w_{\mu \nu} U(\Lambda^{-1})M^{\mu \nu}U(\Lambda) \tag{3}$$
$$\text{ RHS (After substituting $(2)$ in $(1)$) : }I + \frac{i}{2\hbar} {\Lambda^{\mu}}_{\rho}\delta w_{\mu \nu}{\Lambda^{\nu}}_{\sigma} M^{\rho \sigma} \tag{4}$$
Equating $(3)$ and $(4)$, we get $$ \delta w_{\mu \nu} U(\Lambda^{-1})M^{\mu \nu}U(\Lambda)=\delta w_{\mu \nu}{\Lambda^{\mu}}_{\rho}{\Lambda^{\nu}}_{\sigma} M^{\rho \sigma}$$