$42797\; KB = 42797 \times 1024$ bytes require $42797 \times 1024 / 512$ sectors $= 85594$ sectors.
$\langle 1200, 9, 40 \rangle$ is the starting address. So, we can have $24$ sectors in this recording surface. Remaining $85570$ sectors.
$85570$ sectors require $\lceil \frac{85570}{64}\rceil= 1338$ recording surfaces. We start with recording surface $9,$ so we can have $7$ more in the given cylinder. So, we have $1338 - 7 = 1331$ recording surfaces left.
In a cylinder, we have $16$ recording surfaces. So, $1331$ recording surfaces require $\lceil \frac{1331}{16}\rceil = 84$ different cylinders.
The first cylinder (after the current one) starts at $1201.$ So, the last one should be $1284.$
$\langle 1284, 3, 1 \rangle$ will be the end address. $(1331 - 16 \times 83 +1 - 1 = 3$ $(3$ surfaces full and $1$ partial and $-1$ since address starts from $0),$ and $ 85570 - 1337 \times 64 -1 = 1)$