Calculating rate constant of intersystem crossing from Jablonski diagram
The lifetime of the excited state is the reciprocal of the sum of rate constants which you are trying to find:

$$\begin{align}
\tau(T_1) = \frac{1}{\sum_i k_i(T_1)} &amp;= \pu{6 \times 10^-3 s} \\
\sum_i k_i(T_1) &amp;= \frac{1}{\pu{6 \times 10^-3 s}} \\
&amp;= \pu{167 s^-1}
\end{align}$$

The two lines going from the $T_1 \to S_0$ state represent intersystem crossing (ISC) and phosphorescence. Squiggly lines represent nonradiative transitions, of which ISC is one, so the quantum yield for ISC is 10%. (Straight lines represent radiative transitions, in this case phosphorescence.)

Therefore:

$$\begin{align}
k_\text{ISC} &amp;= \Phi_\text{ISC} \cdot \sum_i k_i(T_1) \\
&amp;= 0.1 \cdot \pu{167 s^-1} \\
&amp;= \pu{16.7 s^-1}
\end{align}$$