We can estimate the transition probability by adopting the approximation: the radial wave function is assumed to be constant inside the nucleus and zero outside. Then, by the normalization condition,
\begin{eqnarray} \int g_{nl}(r) g_{nl}(r) r^2 dr \approx g_{nl}^2 \int_0^R r^2 dr = g_{nl}^2 \frac{R^3}{3} = 1 \Rightarrow g_{nl} \approx \sqrt{\frac{3}{R^3}} \ \ {\rm at} \ r \le R. \end{eqnarray}
Using this approximation, we can evaluate the radial integral as follows:
\begin{eqnarray} {\cal R}_{ab}^{(\lambda)} \approx \frac{3}{R^3} \int_0^R r^{\lambda + 2} dr = \frac{3}{\lambda + 3} R^\lambda. \end{eqnarray}
For the stretched case: $j_a = 1/2$ and $j_b = \lambda + 1/2$, we can evalute the $3j$ symbol term:
\begin{eqnarray} \hat{\lambda} \hat{j}_a \hat{j}_b \begin{pmatrix} j_a & j_b & \lambda \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{pmatrix} = (-1)^\lambda \sqrt{2\lambda +1 }, \end{eqnarray}
here we assume that the party factor is unity. As a result, the transition probability can be approximated as follows:
\begin{eqnarray} B(E \lambda) \approx \frac{e^2}{4\pi} \left( \frac{3}{3+\lambda} \right)^2 R^{2\lambda}. \end{eqnarray}
In addition, if we use the relation of $R = r_0 A^{1/3} = 1.2 A^{1/3}\,{\rm fm}$,
\begin{eqnarray} B_W(E \lambda) = \frac{1.2^{2\lambda}}{4\pi} \left( \frac{3}{\lambda + 3} \right)^2 A^{2\lambda /3} e^3 {\rm fm}^{2\lambda}. \end{eqnarray}
This is called Weissfopf single-particle estimate or Weisskopf unit (W.u.).
For the magnetic transition, one can take
\begin{eqnarray} B_W(M\lambda) = \frac{10}{\pi} 1.2^{2\lambda -2} \left( \frac{3}{\lambda +3} \right)^2 A^{(2 \lambda -2)/3} (\mu_N/c)^2 \,{\rm fm}^{2\lambda -2}. \end{eqnarray}
This is a very simple formula to estimate the transition probability because it only depends on the nuclear mass number $A$. In particular, it is useful to estimate the decay half-lives. However, we should be aware that this is only for order-of-magnitude approximations.
