As far as I understand, Average memory access time in this case should be,

$ (0.3 * 2000 + 0.7 * 3000)* \{(1 - \frac{100}{2000}) * 2 + (\frac{100}{2000}) * (1 - \frac{60}{2000}) * 10 + (\frac{100}{2000} * \frac{60}{2000} ) * 200\}$

I am not sure if we should include for example, L1 access time when it has a miss and eventually finds the content in L2 instead?

$ (0.3 * 2000 + 0.7 * 3000)* \{(1 - \frac{100}{2000}) * 2 + (\frac{100}{2000}) * (1 - \frac{60}{2000}) * 10 + (\frac{100}{2000} * \frac{60}{2000} ) * 200\}$

I am not sure if we should include for example, L1 access time when it has a miss and eventually finds the content in L2 instead?

http://gateoverflow.in/tag/cache-memory

Now, try solving this.