Let's consider the group G of order 224. By the Sylow theorems, we know that the number of Sylow 2-subgroups of G, denoted by n2, must divide the order of G and be congruent to 1 modulo 2. Additionally, the number of Sylow 7-subgroups of G, denoted by n7, must divide the order of G and be congruent to 1 modulo 7.

Since the order of G is 224 = 2^5 * 7, the possible values for n2 are 1, 7, 14, 28, 56, or 112. Similarly, the possible values for n7 are 1, 8, 16, or 112.

Now, let's consider the element a in G. Since the order of a must divide the order of G, the possibilities for the order of a are 1, 2, 4, 7, 8, 14, 16, 28, 32, 56, 112, or 224.

We can now analyze the possible values for n2 and n7 along with the possible orders of a:

1) If n2 = 1 and n7 = 1:
In this case, G has a unique Sylow 2-subgroup and a unique Sylow 7-subgroup. Since the order of G is not a prime number, G must have a non-identity element of order 2 or 7. Hence, a cannot be the identity element.

2) If n2 > 1 and n7 > 1:
In this case, G has multiple Sylow 2-subgroups and multiple Sylow 7-subgroups. By the Sylow theorems, the number of Sylow 2-subgroups and the number of Sylow 7-subgroups must divide the order of G. Therefore, a cannot have an order that divides 224.

3) If n2 = 1 and n7 > 1:
In this case, G has a unique Sylow 2-subgroup and multiple Sylow 7-subgroups. Similar to case 1, G must have a non-identity element of order 2 or 7. Hence, a cannot be the identity element.

4) If n2 > 1 and n7 = 1:
In this case, G has multiple Sylow 2-subgroups and a unique Sylow 7-subgroup. By the Sylow theorems, the number of Sylow 2-subgroups and the number of Sylow 7-subgroups must divide the order of G. Therefore, a cannot have an order that divides 224.

Based on the above analysis, we conclude that there is no element a in G such that the order of a divides 224.