I'm afraid I don't know what 'ICE BOX' means in this context: I assume it's a mnemonic device intended to help you remember how to calculate a K_a.
The molar mass, m, of formic acid is 50 gmol^(-1), so the number of moles is given by n=m/M = 23/50 = 0.46 mol.
Dissolving 0.46 mol of formic acid in 10 L of water yields to a concentration of C=n/V = 0.46/10 = 0.046 mol.
In this case, formic acid dissociates as follows:
HCOOH \rightleftharpoons H^+ + HCOO^-
The expression for K_a is as follows:
K_a = ([H^+][HCOO^-])/([HCOOH])
(we can ignore [H_2O] because it is effectively constant)
We are told that [H^+] = 3 \times 10^-3 molL^-1
(molL^-1 is a more technically-correct way than M of writing a concentration in SI units)
Looking at the equation, each mole of formic acid that dissociates yields one mole of H^+ and one mole of HCOO^-. Substituting these values into the K_a expression yields:
K_a = ([H^+][HCOO^-])/([HCOOH]) = ([3 \times 10^-3][3 \times 10^-3])/([0.046 - (3 \times 10^-3)]) = 2.1 \times 10^(-4)
This is a unitless constant.
