Capacity (in bytes) $=10 \times 256 \times 512 \times 1024$
Capacity (in bytes) $=10 \times 2^8 \times 2^9 \times 2^{10}$
Capacity (in bytes) $=10 \times 2^{27} ~\text{bytes}$
Capacity (in Mbyte) = $\mathbf{1 2 8 0} ~\text{Mbyte}$
Bits for Surfaces $(S=10)$ : $\text{Bits}_{\text {Surface }}=\left\lceil\log _2(10)\right\rceil=\mathbf{4} ~\text{bits}$ (since $2^3=8$ is too small, and $2^4=16$ is sufficient).
Bits for Tracks $(T=256)$ : $\text{Bits}_{\text {Track }}=\log _2(256)=\mathbf{8} ~\text{bits}$ (since $2^8=256$ ).
Bits for Sectors $(R=512)$ : $\text{Bits}_{\text {Sector }}=\log _2(512)=\mathbf{9} ~\text{bits}$ (since $2^9=512$ ).
Total Bits $=4+8+9=\mathbf{21} ~\text{bits}$
