The two pointers take $4+4=8$ bytes.
The date requires $10$ bytes, but because of the alignment rule its occupied space is rounded up to the next multiple of $4:10\rightarrow12$
Hence record-header size $:4+4+12=20$ bytes.
Now, The real field needs $6$ bytes $:6\rightarrow8$
The character string needs $17$ bytes $:17\rightarrow20$
The timestamp already occupies a multiple of $4:8\rightarrow8$
Therefore, field storage is:
$8+20+8=36$ bytes.
Total physical record size:
$20+36=56$ bytes.
Now, There are eight $4$-byte integers:
$8\times4=32$ bytes.
So the space available for records is:
$4096-32=4064$ bytes.
Now,
$\left\lfloor\dfrac{4064}{56}\right\rfloor=\boxed{72}$
