Metal `M` of radius `50 nm` is crystallized in fcc type and made cubical crystal such that face of unit cells aligned with face of cubical crystal. If the total number of metal atoms of `M` at all faces of cubical crystal is `6 xx 10^(30)`, then the area of one face of cubical crystal is `A xx 10^(16) m^(2)`. Find the value of `A`.

Explanation

To solve the problem step by step, we will follow the logical flow of the information given in the question.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a metal `M` with a radius of `50 nm` that crystallizes in a face-centered cubic (FCC) structure.
- The total number of metal atoms at all faces of the cubic crystal is given as `6 x 10^30`.

2. **Calculating Number of Atoms on One Face:**
- A cube has `6 faces`. Therefore, the number of atoms on one face can be calculated as:
\[ \text{Number of atoms on one face} = \frac{6 \times 10^{30}}{6} = 1 \times 10^{30} \]

3. **Finding the Side Length of the Cube:**
- In an FCC structure, the relationship between the radius `r` of the atom and the side length `a` of the unit cell is given by:
\[ a = 4r \]
- Given the radius `r = 50 nm = 50 \times 10^{-9} m`, we can calculate:
\[ a = 4 \times (50 \times 10^{-9}) = 200 \times 10^{-9} m = 2 \times 10^{-7} m \]

4. **Calculating the Area of One Face of the Cube:**
- The area `A` of one face of the cube is given by:
\[ A = a^2 \]
- Substituting the value of `a`:
\[ A = (2 \times 10^{-7})^2 = 4 \times 10^{-14} m^2 \]

5. **Expressing the Area in the Required Form:**
- The problem states that the area of one face of the cubic crystal is `A x 10^{16} m^2`. We need to express `4 x 10^{-14} m^2` in this form:
\[ 4 \times 10^{-14} = A \times 10^{16} \]
- Rearranging gives:
\[ A = 4 \times 10^{-14} \times 10^{-16} = 4 \times 10^{-30} \]

6. **Final Calculation of A:**
- Since we need to find the value of `A` such that the area is expressed as `A x 10^{16}`, we can see that:
\[ A = 4 \]

### Final Answer:
The value of `A` is `4`.