(N/A) Let the dimensions of the cuboidal box be $6x$,$4x$,and $3x$ meters.
The total surface area of the cuboid is $SA = 2(lb + bh + lh) = 2((6x)(4x) + (4x)(3x) + (6x)(3x)) = 2(24x^2 + 12x^2 + 18x^2) = 2(54x^2) = 108x^2 \, m^2$.
The cost of covering with white paper at $₹ 2.00/m^2$ is $2 \times 108x^2 = 216x^2$.
The cost of covering with coloured paper at $₹ 2.50/m^2$ is $2.50 \times 108x^2 = 270x^2$.
The difference in cost is $270x^2 - 216x^2 = 54x^2$.
Given that the difference is $₹ 1350$,we have $54x^2 = 1350$.
$x^2 = 1350 / 54 = 25$.
$x = 5$.
Thus,the dimensions are $6(5) = 30 \, m$,$4(5) = 20 \, m$,and $3(5) = 15 \, m$.