Calculating rate constant of intersystem crossing from Jablonski diagram

I am to calculate the rate constant for $T\_1 \to S\_0$ intersystem crossing, with reference to the Jablonski diagram attached below. I know this should not be difficult, but for some reason I cannot put the number together to get the correct result.


I have that


$$
\Phi\_\text{ISC} = \frac{k\_\text{ISC}}{\sum\_i k\_i}
$$


where the denominator represents a sum of all decay pathways for the $T\_1$ state. However, how can this sum be calculated if we do not know the rate constants to begin with?


[![enter image description here](https://i.stack.imgur.com/TD57S.png)](https://i.stack.imgur.com/TD57S.png)




The lifetime of the excited state is the reciprocal of the sum of rate constants which you are trying to find:


$$\begin{align}
\tau(T\_1) = \frac{1}{\sum\_i k\_i(T\_1)} &= \pu{6 \times 10^-3 s} \\
\sum\_i k\_i(T\_1) &= \frac{1}{\pu{6 \times 10^-3 s}} \\
&= \pu{167 s^-1}
\end{align}$$


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:


$$\begin{align}
k\_\text{ISC} &= \Phi\_\text{ISC} \cdot \sum\_i k\_i(T\_1) \\
&= 0.1 \cdot \pu{167 s^-1} \\
&= \pu{16.7 s^-1}
\end{align}$$


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