Let the order of the B+ tree be $n$.

$n\times(\text{block pointer})+(n-1)\times(\text{key pointer}) \leq \text{block size}$

$8n + 12(n-1) \leq 4096$

$20n – 12 \leq 4096$

$20n \leq 4108$

$n \leq 205.4$

$n = 205$

Now all the actual data elements are stored in the leaves of the B+ tree, as we have 1 million ($10^6$) records, the level number of a B+ tree with order 205 that can hold greater than or equal to 1 million elements can be found using

$204^x \geq 10^6 \implies x = 3$

So 3 disk accesses?