file system inode

Question

Consider a UNIX-like file system implemented with i-nodes that resides on a disk of size 512GB. Each i-node has a total of 15 block addresses consisting of direct and indirect block addresses.

suppose the implementation wants to support file size upto 1GB using only direct block address and single indirect block address.Atleast how many of the 15 blocks addresses should be used as single indirect block address

a) 0

b) 1

c) 2

d) 3

Comments

Going with the flow...

$1GB$ means  $2^{30} \;bits$

Assuming a single block is of size $4 KB$ i.e. $2^{12} bits$

Total blocks: $\frac{2^{30}}{2^{12}} = 2^{18}$

A single inode gives $15$ block addresses,

if we use the last one (1 block) for indirect addressing that will give: 14 + $ \frac{2^{12}}{2^{5}} $ (assuming 32 bit addresses) $\approx 2^7$ addresses.

so basically using a single block for single indirect addressing gives $2^7$ block addresses, so solving for $x$ in:

$x * 2^7 = 2^{18}$, which will give $x = 2^{11}$, which is $ \textbf{Wrong} $

Anyone care to correct what happened wrong?